$3y^2+x^2-xy=1$ Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1-2x}{6y+1}$ (Choice B) B $\dfrac{y-2x}{6y+x}$ (Choice C) C $\dfrac{y-2x}{6y-x}$ (Choice D) D $\dfrac{1-2x}{6y-1}$
Solution: We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $3y^2+x^2-xy=1$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} 3y^2+x^2-xy&=1 \\\\ \dfrac{d}{dx}(3y^2+x^2-xy)&=\dfrac{d}{dx}(1) \\\\ \dfrac{d}{dx}(3y^2)+\dfrac{d}{dx}(x^2)-\dfrac{d}{dx}(xy)&=0 \\\\ 6y\cdot\dfrac{dy}{dx}+2x-\Bigl(1\cdot y+x\cdot\dfrac{dy}{dx}\Bigr)&=0 \\\\ 6y\cdot\dfrac{dy}{dx}+2x-y-x\cdot\dfrac{dy}{dx}&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 6y\cdot\dfrac{dy}{dx}+2x-y-x\cdot\dfrac{dy}{dx}&=0 \\\\ \dfrac{dy}{dx}(6y-x)&=y-2x \\\\ \dfrac{dy}{dx}&=\dfrac{y-2x}{6y-x} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{y-2x}{6y-x}$.